M & Ns... Not appealing anymore?

This seems incorrect. The answer should be $8$ . If you restricted to $n \le m$ or nonnegative integers $n$ , then $9$ would be correct.

$2007^{2008} = 3^{4016} \cdot 223^{2008}$ is the prime factorization. But it's clear that only one of the three factors can be divisible by $3$ and only one can be divisible by $223$ , so one of the factors $n, 2n+1, 5n+2$ has to be divisible by $3^{4016}$ and one of them has to be divisible by $223^{2008}$ . This is necessary and sufficient for the product to be divisible by $m$ .

So this is a Chinese Remainder Theorem computation. There are $9$ total solutions mod $m$ , since there are $3$ choices for the factor in both conditions, but one of the solutions is $0$ , which is ruled out by the problem. So the correct answer is $\fbox{8}$ .